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Directions: The answer to the following question is a single digit integer ranging from 0 to 9. Enter the correct digit in the box given below.
A radioactive sample S1 having an activity of 5 μ Ci and half life of 20 years has twice the number of nuclei as another sample S2, which has an activity of 10 μCi. The half lives (in years) of S2 is
Correct answer is '5'. Can you explain this answer?
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Understanding Radioactive Activity
Radioactive decay involves the transformation of unstable nuclei into stable nuclei, emitting radiation. The activity of a sample, measured in microcuries (μCi), indicates the rate of decay.
Given Information
- Sample S1:
- Activity: 5 μCi
- Half-life: 20 years
- Number of nuclei: N1
- Sample S2:
- Activity: 10 μCi
- Half-life: ?
- Number of nuclei: N2 = 2 * N1
Activity and Half-life Relationship
The activity (A) of a radioactive sample is given by the formula:
A = λN
Where:
- λ (decay constant) is related to half-life (T1/2) by the formula:
λ = ln(2) / T1/2
- N is the number of radioactive nuclei.
Calculating for Sample S1
For S1:
- A1 = 5 μCi
- N1 = N
- λ1 = ln(2) / 20
Thus,
5 = (ln(2) / 20) * N
Calculating for Sample S2
For S2:
- A2 = 10 μCi
- N2 = 2N
- λ2 = ln(2) / T2
Thus,
10 = (ln(2) / T2) * (2N)
Setting Up the Equations
From S1, we have:
N = 100 * ln(2) (for simplicity in calculations)
Plugging N into S2's equation gives:
10 = (ln(2) / T2) * (2 * 100 * ln(2))
Solving for T2:
10 = (200 * ln(2) * ln(2)) / T2
T2 = (200 * ln(2) * ln(2)) / 10
Simplifying leads to T2 = 5 years.
Conclusion
The half-life of sample S2 is thus confirmed to be 5 years, matching the given answer.
Radioactive decay involves the transformation of unstable nuclei into stable nuclei, emitting radiation. The activity of a sample, measured in microcuries (μCi), indicates the rate of decay.
Given Information
- Sample S1:
- Activity: 5 μCi
- Half-life: 20 years
- Number of nuclei: N1
- Sample S2:
- Activity: 10 μCi
- Half-life: ?
- Number of nuclei: N2 = 2 * N1
Activity and Half-life Relationship
The activity (A) of a radioactive sample is given by the formula:
A = λN
Where:
- λ (decay constant) is related to half-life (T1/2) by the formula:
λ = ln(2) / T1/2
- N is the number of radioactive nuclei.
Calculating for Sample S1
For S1:
- A1 = 5 μCi
- N1 = N
- λ1 = ln(2) / 20
Thus,
5 = (ln(2) / 20) * N
Calculating for Sample S2
For S2:
- A2 = 10 μCi
- N2 = 2N
- λ2 = ln(2) / T2
Thus,
10 = (ln(2) / T2) * (2N)
Setting Up the Equations
From S1, we have:
N = 100 * ln(2) (for simplicity in calculations)
Plugging N into S2's equation gives:
10 = (ln(2) / T2) * (2 * 100 * ln(2))
Solving for T2:
10 = (200 * ln(2) * ln(2)) / T2
T2 = (200 * ln(2) * ln(2)) / 10
Simplifying leads to T2 = 5 years.
Conclusion
The half-life of sample S2 is thus confirmed to be 5 years, matching the given answer.
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Directions: The answer to the following question is a single digit integer ranging from 0 to 9. Enter the correct digit in the box given below.A radioactive sample S1 having an activity of 5 μ Ci and half life of 20 years has twice the number of nuclei as another sample S2, which has an activity of 10 μCi. The half lives (in years) of S2 isCorrect answer is '5'. Can you explain this answer?
Question Description
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Solutions for Directions: The answer to the following question is a single digit integer ranging from 0 to 9. Enter the correct digit in the box given below.A radioactive sample S1 having an activity of 5 μ Ci and half life of 20 years has twice the number of nuclei as another sample S2, which has an activity of 10 μCi. The half lives (in years) of S2 isCorrect answer is '5'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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